Why is no qualification necessary?

Question

OK, I\'ll just post the complete program even though it has extraneous stuff and the code in question is the dead code…

#include 
#include         


        

Answer 1

The difference is that ifstream isn't visible as an injected class name because it is the name of a typedef, not the name of the class. It isn't therefore visible unqualified as an injected class name from the base class.

ios_base is a genuine class name which is a base class (of a base class) of the class where it is used and so is visible unqualified as an inject class name.

E.g.

namespace X
{
    class A {};
    template class Z {};
    typedef Z B;
}

class C : public X::A
{
    C() : A() {} // OK, A is visible from the base class
};

class D : public X::B
{
    D() : B() {} // Error, B is a typedef,
    // : X::B(), : Z() or even : Z() can be used.
};

In your example, instead of std::ifstream, you can use unqualified basic_ifstream instead. (Or basic_ifstream or basic_ifstream > but these don't really save any typing or help clarity at all.)